Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
int main()
{
	int n;
	cin>>n;
	double p[n];
	int i,j;
	for(i=0;i<n;i++)
	cin>>p[i];
	double dp[n+1][n+1];
	for(i=0;i<=n;i++)
	dp[0][i]=0.0;
	dp[0][0]=1.0;
	for(i=1;i<=n;i++)
	dp[i][0]=dp[i-1][0]*(1-p[i-1]);
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=n;j++)
			dp[i][j]=((1-p[i-1])*dp[i-1][j])+(p[i-1]*dp[i-1][j-1]);
	}
	double sum=0;
	for(i=(n+1)/2;i<=n;i++)
		sum=sum+dp[n][i];
	cout<<setprecision(9)<<sum;
}